From the Isometry to Moments
For deterministic integrands, the Ito isometry gives a direct route from the size of an integrand to the variance of its integral. The same principle applies when the integrand is random and adapted: the squared integrand is averaged over both time and probability. Polarization then turns the isometry into a covariance formula for pairs of integrals. These formulas require square integrability; they do not require the integrands to be deterministic or independent of Brownian motion.
Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) whose increments after time \(s\) are independent of \(\mathcal{F}_s\). Fix a finite \(T>0\). Every integral below is an Ito integral on \([0,T]\), with a progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); this class is denoted by \(\mathcal{L}^2\). For two such integrands \(H,G\), write \(I_T(H)=\int_0^T H_s\,dB_s\) and \(I_T(G)=\int_0^T G_s\,dB_s\). Both terminal integrals are square-integrable random variables by the Ito isometry.
The variance formula is the Ito isometry together with the zero-mean property. The covariance formula says that the relevant inner product is the expected time-integrated product of the integrands, not the product of their separate expectations. Here \(\operatorname{Cov}(X,Y)=\mathbb{E}[(X-\mathbb{E}X)(Y-\mathbb{E}Y)]\) for square-integrable random variables \(X,Y\).
Why the Formulas Hold
The Ito isometry, established for the Ito integral, states that for each \(K\in\mathcal{L}^2\), \(\mathbb{E}[I_T(K)^2]=\mathbb{E}\int_0^T K_s^2\,ds\). We first justify the zero mean, then use this isometry to obtain the covariance.
Proof. For a simple predictable integrand \(K_s=\sum_{i=1}^n \xi_i\mathbf{1}_{(t_{i-1},t_i]}(s)\), with \(\xi_i\) square-integrable and \(\mathcal{F}_{t_{i-1}}\)-measurable, its integral is \(\sum_i\xi_i(B_{t_i}-B_{t_{i-1}})\). Each summand is integrable by Cauchy–Schwarz. Independence of the Brownian increment from \(\mathcal{F}_{t_{i-1}}\), and its zero mean, give \(\mathbb{E}[\xi_i(B_{t_i}-B_{t_{i-1}})]=0\). Thus the simple integral has mean zero. For general \(K\in\mathcal{L}^2\), simple predictable integrands approximate \(K\) in the norm \((\mathbb{E}\int_0^T K_s^2\,ds)^{1/2}\), and the Ito isometry implies convergence of their integrals in \(L^2(\mathbb{P})\). Since \(L^2\) convergence implies \(L^1\) convergence on a probability space, their expectations converge to \(\mathbb{E}[I_T(K)]\); hence that expectation is zero. The isometry now gives \(\operatorname{Var}(I_T(H))=\mathbb{E}[I_T(H)^2]=\mathbb{E}\int_0^T H_s^2\,ds\). Finally, \(H+G\) and \(H-G\) are in \(\mathcal{L}^2\), and polarization gives \[ \mathbb{E}[I_T(H)I_T(G)] =\frac14\left(\mathbb{E}[I_T(H+G)^2]-\mathbb{E}[I_T(H-G)^2]\right) =\mathbb{E}\int_0^T H_sG_s\,ds. \] The zero means make this expectation the covariance. The product on the right is integrable because \(\mathbb{E}\int_0^T |H_sG_s|\,ds\le (\mathbb{E}\int_0^T H_s^2\,ds)^{1/2} (\mathbb{E}\int_0^T G_s^2\,ds)^{1/2}\). This also verifies finiteness. \(\square\)
The inner product formula has two useful consequences. First, Cauchy–Schwarz gives \(\left|\operatorname{Cov}(I_T(H),I_T(G))\right| \le \sqrt{\operatorname{Var}(I_T(H))\operatorname{Var}(I_T(G))}\). Second, if \(H\) and \(G\) are orthogonal in the space with inner product \(\mathbb{E}\int_0^T H_sG_s\,ds\), their integrals are uncorrelated. Zero covariance does not, by itself, imply independence. The variance formula also handles degeneracy: \(I_T(H)\) has variance zero precisely when \(H=0\) almost everywhere with respect to the product of \(\mathbb{P}\) and time measure. In that case its centered integral is zero almost surely.
Worked Example: The Integral of Brownian Motion
On \([0,2]\), take \(H_s=B_s\). This is an admissible integrand since it is progressively measurable and \(\mathbb{E}\int_0^2 B_s^2\,ds=\int_0^2 s\,ds=2<\infty\). Find the mean and variance of \(\int_0^2 B_s\,dB_s\).
The theorem gives mean zero. Since \(\mathbb{E}[B_s^2]=s\), the variance is \(\int_0^2 s\,ds=[s^2/2]_0^2=2\). For an independent check, apply Ito's formula to \(f(x)=x^2\), for which \(f'(x)=2x\) and \(f''(x)=2\). The correction term is \(\tfrac12 f''(B_s)\,ds=ds\), so \(d(B_s^2)=2B_s\,dB_s+ds\), and consequently \(\int_0^2 B_s\,dB_s=(B_2^2-2)/2\).
Here \(B_2\sim N(0,2)\), so \(\mathbb{E}[B_2^2]=2\) and \(\mathbb{E}[B_2^4]=3(2^2)=12\). The displayed expression has mean \((2-2)/2=0\). Its variance is \(\mathbb{E}[(B_2^2-2)^2]/4=(12-4\cdot2+4)/4=8/4=2\), agreeing with the isometry.
Covariance and Orthogonal Integrands
Orthogonality concerns the expected product integrated over time. The integrands may be random and may be nonzero at the same times. The next examples contrast this with disjoint time intervals and show how covariance can be checked directly as well as by polarization.
Worked Example: Zero Covariance with Overlapping Integrands
On \([0,1]\), let \(H_s=1\) and \(G_s=B_s\). Both are in \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^1 1\,ds=1\) and \(\mathbb{E}\int_0^1 B_s^2\,ds=\int_0^1s\,ds=1/2\). Calculate the covariance and both variances.
The covariance formula gives \(\operatorname{Cov}(I_1(H),I_1(G))=\int_0^1\mathbb{E}[B_s]\,ds=0\). The variances are \(\operatorname{Var}(I_1(H))=\int_0^1 1\,ds=1\) and \(\operatorname{Var}(I_1(G))=\int_0^1s\,ds=1/2\). For a direct check, \(I_1(H)=B_1\). Ito's formula for \(f(x)=x^2\), with \(f'(x)=2x\), \(f''(x)=2\), and correction \(\tfrac12 f''(B_s)\,ds=ds\), gives \(I_1(G)=(B_1^2-1)/2\). Thus \(\mathbb{E}[B_1(B_1^2-1)/2]=(\mathbb{E}[B_1^3]-\mathbb{E}[B_1])/2=0\), since \(B_1\) is centered normal. Also, \(\operatorname{Var}((B_1^2-1)/2)=(3-2+1)/4=1/2\), using \(\mathbb{E}[B_1^4]=3\) and \(\mathbb{E}[B_1^2]=1\). This verifies both results. The integrands overlap in time, but their expected product is zero.
Worked Example: A Nonzero Covariance from a Random Integrand
On \([0,1]\), take \(H_s=1\) and \(G_s=B_s^2\). The second integrand is admissible because \(\mathbb{E}\int_0^1 B_s^4\,ds=\int_0^1 3s^2\,ds=1\). Find the covariance and variances.
By the theorem, \(\operatorname{Cov}(I_1(H),I_1(G))=\int_0^1\mathbb{E}[B_s^2]\,ds=\int_0^1s\,ds=1/2\). The variances are \(\operatorname{Var}(I_1(H))=1\) and \(\operatorname{Var}(I_1(G))=\int_0^1\mathbb{E}[B_s^4]\,ds=\int_0^1 3s^2\,ds=1\). For a direct covariance check, apply Ito's formula to \(f(x)=x^3\): \(f'(x)=3x^2\), \(f''(x)=6x\), and the correction is \(\tfrac12 f''(B_s)\,ds=3B_s\,ds\). Therefore \(I_1(G)=B_1^3/3-\int_0^1 B_s\,ds\). Since \(I_1(H)=B_1\), the covariance is \(\mathbb{E}[B_1^4]/3-\int_0^1\mathbb{E}[B_1B_s]\,ds\). For \(s\le1\), \(\mathbb{E}[B_1B_s]=s\), so this equals \(3/3-\int_0^1s\,ds=1-1/2=1/2\), as required. The theorem computes the covariance from the expected integrand product, even though the integrand \(B_s^2\) is random.
Worked Example: Orthogonality from Disjoint Time Supports
On \([0,2]\), let \(H_s=\mathbf{1}_{(0,1]}(s)\) and \(G_s=\mathbf{1}_{(1,2]}(s)B_1\). The second integrand is adapted on its interval and square-integrable because \(\mathbb{E}\int_1^2 B_1^2\,ds=\mathbb{E}[B_1^2](2-1)=1\). Compute the covariance and variances.
At every time except the single endpoint \(s=1\), at least one integrand is zero. Thus \(\mathbb{E}\int_0^2H_sG_s\,ds=0\), and the covariance is zero. The first variance is \(\int_0^1 1\,ds=1\). The second is \(\mathbb{E}\int_1^2B_1^2\,ds=1\). To check, the integrals are \(B_1\) and \(B_1(B_2-B_1)\). Their covariance is \(\mathbb{E}[B_1^2(B_2-B_1)]\). The increment \(B_2-B_1\) is independent of \(B_1\) and has mean zero, so this expectation is zero. The second integral has second moment \(\mathbb{E}[B_1^2]\mathbb{E}[(B_2-B_1)^2]=1\cdot1=1\), agreeing with its variance formula.
Common Mistakes and Practical Meaning
A frequent error is to replace the expected square of a random integrand by the square of its expectation. For example, on \([0,1]\), \(H_s=B_s\) has \(\mathbb{E}[H_s]=0\) at every time. The incorrect calculation \(\int_0^1(\mathbb{E}B_s)^2\,ds=0\) would claim that \(\int_0^1B_s\,dB_s\) has variance zero. The correct isometry calculation is \(\mathbb{E}\int_0^1B_s^2\,ds=\int_0^1s\,ds=1/2\). The square must be taken before expectation.
Likewise, \(\mathbb{E}[H_sG_s]\) cannot generally be replaced by \(\mathbb{E}[H_s]\mathbb{E}[G_s]\): random integrands can be dependent. The covariance theorem uses the product at the same time and then averages it. This is useful whenever a model has several Ito integrals driven by the same Brownian motion: expected integrated products reveal whether their terminal values are correlated and quantify that correlation. The formula is for Ito integrals under \(\mathbb{P}\) on the stated filtered space; no change of measure or Stratonovich correction is involved.
Check Your Understanding
Use the isometry and covariance formula, stating the integrability condition for each integrand.
- For \(H_s=2\) on \([0,3]\), find the mean and variance of \(\int_0^3 H_s\,dB_s\).
- For \(H_s=B_s\) on \([0,1]\) and \(G_s=1\), compute the covariance and explain why it does not imply independence.
- On \([0,1]\), take \(H_s=B_s\) and \(G_s=B_s\). Find the covariance and identify the integrand inner product that gives it.
- Give a condition on \(H\) and \(G\) that guarantees their Ito integrals have zero covariance, and distinguish this from independence.
- If \(\mathbb{E}\int_0^T H_s^2\,ds=0\), what does the variance formula imply about \(I_T(H)\)?